Please guys, help me out on this.
Find the extremals of
I = [x1,x2] F(x, y, y')dx
F= (y')^2 +2y
Please, or just hint and show the steps and formulas. Cuz, I'm really confused. (((
The E–L equation says that the condition for an extremal is $\displaystyle \textstyle \frac{\partial F}{\partial y} - \frac d{dx}\frac{\partial F}{\partial y'} = 0$.
If $\displaystyle F(x,y,y') = y'^2+2y$ then $\displaystyle \tfrac{\partial F}{\partial y} = 2$ and $\displaystyle \tfrac{\partial F}{\partial y'} = 2y'$. So the E–L equation says that $\displaystyle 2-(2y')' = 0$, or $\displaystyle y''=1$.
Now you have a simple differential equation for y, whose general solution is $\displaystyle y = \tfrac12x^2+ax+b$. The constants a and b would have to be determined by some initial conditions for the problem (such as the values of y when x=x_1 and x= x_2).
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Use the same procedure that I showed for your similar question in the Calculus section (which is where this thread also belongs, Moderator!). What is $\displaystyle \tfrac{\partial F}{\partial y}$? What is $\displaystyle \tfrac{\partial F}{\partial y'}$? What does the Euler–Lagrange equation say? Can you solve it?
You need to show that you have made some attempt at progress before you get any more help on this one.
This thread has got inextricably tangled up with the other one on this topic (maybe my fault for suggesting that it should be moved). I think that we're now supposed to be looking at the extremals associated with the function $\displaystyle F(x,y,y') = y'^2-ky^2$. I have no idea where the equations F - y'F_y' = c and y' = +- (-(ky)^2 - c)^1/2 come from.
What you should do is to find the partial derivatives $\displaystyle \tfrac{\partial F}{\partial y}$ and $\displaystyle \tfrac{\partial F}{\partial y'}$, and plug them in to the Euler–Lagrange equation $\displaystyle \textstyle \frac{\partial F}{\partial y} - \frac d{dx}\frac{\partial F}{\partial y'} = 0$. You then get a differential equation for y. For this example, the equation should be $\displaystyle y''+ky=0$, and the solutions will be trigonometric or exponential functions, depending on whether k is positive or negative.